I don't know if this is a well know fact but I have observed that every number no matter how large that is equally divided by $9$, will equal $9$ if you add all the numbers it is made from until there is $1$ digit.

$\begingroup$Here's something else to try: write a number $n$ such that the digits, read left to right, are increasing (e.g. 12345 or 13579). What is the sum of the digits of $9 \times n$?$\endgroup$
– ElliottDec 31 '10 at 20:32

6 Answers
6

Not quite right, since $9\times 0 = 0$ and the digits don't add up to $9$; but otherwise correct.

The reason it works is that we write numbers in base $10$, and when you divide $10$ by $9$, the remainder is $1$. Take a number, say, $184631$ (I just made it up). Remember what that really means:
$$184631 = 1 + 3\times 10 + 6\times 10^2 + 4\times 10^3 + 8\times 10^4 + 1\times 10^5.$$
The remainder when you divide any power of $10$ by $9$ is again just $1$, so adding the digits gives you a number that has the same remainder when dividing by $9$ as the original number does. Keep doing it until you get down to a single digit and you get the remainder of the original number when you divide by $9$, except that you get $9$ instead of $0$ if the number is a nonzero multiple of $9$.

But since every multiple of $9$ is a multiple of $9$, you will always get $9$.

Note that you have a similar phenomenon with $3$ (a divisor of $9$), since adding the digits of a multiple of $3$ will always result in one of the one-digit multiples of $3$: $3$, $6$, or $9$.

If we wrote in base $8$, instead of base $10$, then $7$ would have the property: if you write a number in base $8$ and add the digits (in base 8) until you get down to a single digit between $1$ and $7$, then the multiples of $7$ will always end yielding $7$, for precisely the same reason. And if we wrote in base $16$, then $15$ (or rather, F) would have the property. In general, if you write in base $b$, then $b-1$ has the property.

Coda. This reminds me of an anecdote a professor of mine used to relate: a student once came to him telling him he had discovered a very easy way to test divisibility of any number $N$ by any number $b$: write $N$ in base $b$, and see if the last digit is $0$. I guess, equivalently, you could write $N$ in base $b+1$, and add the digits to see if you get $b$ at the end.

$\begingroup$Yes. and another test is to write the number in base $b-1$ and check if the alternating sum adds up to zero. (mimicking the divisibility test for 11 in decimal system)$\endgroup$
– user17762Dec 31 '10 at 20:10

Yours is special case $\rm\ N\equiv 0\pmod{\! 9}\ $ so the above implies that it remains $\equiv 0\ $ when you map $\rm N$ to its digit sum $\rm\:S_N.\:$ Since this map is strictly decreasing for $\rm\:N > 9,\ $ iterating it must eventually reach some nonzero $\rm\: N' \le 9.\ $ But $\rm\ N'\equiv 0\pmod{\!9}\, $ so we conclude that $\rm\: N' = 9.$

Note: if modular arithmetic is unfamiliar then one may proceed more simply as follows:

The analogous result holds true for any radix $\rm\:b,\,$ i.e. one can cast $\rm\:b\!-\!1$'s in the same way, since $\rm\ P(b) \equiv P(1)\pmod{\! b\!-\!1}.\:$ Hence $\:9\:$is "special" because it is one less than the radix$10.\,$ Similarly we may cast $11$'s using $\rm\ P(10)\equiv P(-1)\ \ (mod\ 11).\:$ One can devise countless divisibility tests by using modular reduction in this way, e.g. see here for casting out $91$'s.

It deserves to be better known that one may also cast out nines to check rational arithmetic - as long as the fractions have denominator coprime to $3$, e.g. see Hilton; Pedersen: Casting out nines revisited (be aware these results are very old). Analogous remarks hold true for any ring that has $\ \mathbb Z/9\ $ as an image - just as one can apply parity arguments in any ring that has $\ \mathbb Z/2\ $ as an image, e.g. the ring of all rationals with odd denominator, or the ring of Gaussian integers $\,\mathbb Z[i],\,$ where the image $\, \mathbb Z[i]/(2,i\!-\!1) \cong \mathbb Z/2\ $ yields the natural parity definition: $\, a+b\:i\ $ is even $\iff a\equiv b\pmod{\! 2},\,$ i.e. if $\, a+b\:i\ $ maps to $\:0\:$ via the above isomorphism, which maps $\, 2\to 0,\ i\to 1\:$. See here for further discussion of parity in rings of algebraic integers, including examples of number rings with no parity structure, and with more than one parity structure. See also this post for "casting out orders" in cyclic groups, and see this thread for an in-depth comparison of various elementary inductive proofs of casting nines.

These are elementary prototypical examples of problem-solving by way of modular reduction - one of the keystones of abstract algebra. As such one should be sure to understand these simple instances before moving on to more advanced manifestations of modular reduction.

Many ways to say this I suppose - what made sense to me when I first was confronted with this fact is that the number "$abcd$" is $1000a+100b+10c+d=(999a+99b+9c)+a+b+c+d$. The first term is clearly divisible by $9$ so if the digit sum is then "$abcd$" is as well...

This follows from the divisibility test for $9$ and from the fact that the sum of the digits of a natural number with more than one digit is strictly less than the number itself i.e. if $n$ is a natural number with more than one digit and if $s(n)$ denotes the sum of the digits of the natural number then $s(n)<n$. Your observation can be strengthened to note that if we keep repeatedly adding the digits, the final number we are left with is the remainder when divided by $9$.

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